Q. Two persons P and Q crosses the river starting from point A on one side to exactly opposite point B on the other bank of the river. The person P crosses the river in the shortest path. The person Q crosses the river in shortest time and walks back to point B. Velocity of river is 3 kmph and speed of each person is 5 kmph w.r.t river. If the two persons reach the point B in the same time, then the speed of walk of Q is.

(a) 13 *kmph*

(b) 12 *kmph*

(c) 14 *kmph*

(d) 11 *kmph*

**Click to See Answer : **

Sol: Let d = width of river

Time taken to cross the river in shortest path by person P is

$\displaystyle t_p = \frac{d}{\sqrt{5^2-3^2}} = \frac{d}{4}$

Time taken to cross the river in shortest time by person Q is

$ \displaystyle t_Q = \frac{d}{5}$

According to question

$ \displaystyle t_p = t_Q + \Delta t $ ; Where Δt = time taken by person to come to the point B.

If x = walking distance ,

$\displaystyle \Delta t = \frac{x}{v_{walk}} $ ; Where v_{walk} = Walking Speed of man Q

$ \displaystyle \frac{d}{4} = \frac{d}{5} + (v_r\frac{d}{v_Q})\frac{1}{v_{walk}} $

$ \displaystyle \frac{1}{4} = \frac{1}{5} + \frac{3}{5}\frac{1}{v_{walk}} $

$ \displaystyle \frac{1}{4}-\frac{1}{5} = \frac{3}{5}\frac{1}{v_{walk}} $

$ \displaystyle \frac{1}{20} = \frac{3}{5}\frac{1}{v_{walk}} $

$\displaystyle \frac{1}{4} = \frac{3}{v_{walk}} $

v_{walk} = 12 kmph